Twin Paradox (3 objects version)

[h1a8]

TL;DR Summary

I been reading and watching videos on the subject for awhile. Many explain away the paradox in different ways (I don't quite understand some explanations and don't agree on others) but none actually attempt (as far as I know) to actually calculate what both clocks will read in the end (using equations). I included the original word document since the tool may skew the format.

Consider 3 objects, A, B, and C, in relative motion along the x direction.

EVENT 1
B passes A while moving at a constant v = 3/5c relative to A. Both clocks set to t = 0.

EVENT 2
C passes B when B’s clock reads t = 5 while C is moving at a constant v = 3/5c towards (and relative) to A.
C’s clock set to t = 5.

EVENT 3
C passes A.

Illustration

Event 3
< ------------- C <----------------- C <--------------
A
-------------> B -----------------> B ----------------- >
Event 1 Event 2
t_A=t_B=0 t_B=t_C=5

QUESTIONS:
1. What will A’s clock read at Event 2?
2. What will A’s clock read at Event 3?
Here’s my attempt

1. Since B is moving at v_B = 3/5c relative to A then t_A=5/4t_B. So, when t_B=5 then t_A=6.25.
2. Since C is moving at v_C = 3/5c relative to A then t_A=5/4t_C. So, when t_C=10 then t_A=12.5.
Or
2. Since A is moving at v_A = 3/5c relative to C then t_C=5/4t_A. So, when t_C=10 then t_A=8.
Contradiction?

 

[h1a8]

TWIN PARADOX
Consider 3 objects, A, B, and C, in motion along the x direction.​

EVENT 1

B passes A while moving at a constant v = 3/5c relative to A. Both clocks set to t = 0.



EVENT 2

C passes B when B’s clock reads t = 5 while C is moving at a constant v = 3/5c towards (and relative) to A.

C’s clock set to t = 5.



EVENT 3

C passes A.

Illustration

​

Event 3

< ------------- C <----------------- C <--------------

A

-------------> B -----------------> B ----------------- >

Event 1 Event 2

QUESTIONS:​

• What will A’s clock read at Event 2?
• What will A’s clock read at Event 3?

My attempt is in the word document (included).

 

[PeterDonis]

My attempt is in the word document (included).

Word documents, and attachments generally, are not acceptable. You need to post your equations directly in the thread using the PF LaTeX feature. You will find a "LaTeX Guide" link at the bottom left of the post window.

 

[PeterDonis]

I been reading and watching videos on the subject for awhile.

If you have not read the Usenet Physics FAQ article, I strongly suggest that you do so:

https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

Your version does not introduce any meaningful changes from the "instantaneous turnaround" version given in the article (just substitute B for Stella's outbound leg and C for Stella's inbound leg). Needless to say, there is no paradox, and if you are getting different answers doing the calculation different ways, you are making a mistake somewhere. The most likely mistake is that you are forgetting to take into account relativity of simultaneity.

 

[Ibix]

Question 1 is ill-posed and has no unique answer. Not understanding this is why you have different answers to question 2. Your first attempt at question 2 is correct, and you failed to account for the relativity of simultaneity in your second attempt. (You probably failed to account for it in your first attempt too, but happened to pick an approach where it has no effect.)

This kind of simple problem can always be analysed by picking an inertial frame (I'd suggest A's rest frame), writing down the coordinates of interesting events in that frame, then using the Lorentz transforms to calculate times and places in other frames. Don't blindly apply the time dilation equation and hope - it's a special case of the Lorentz transforms and practice is needed to see when you can use it.

In this particular case it would be helpful to use the Lorentz transforms to work out what time A's clock shows at the same time as B and C pass in the rest frames of A, B, and C. Have a go - post your working if you get stuck or need help interpreting the answers.

Probably the more helpful solution is to draw spacetime diagrams (also called Minkowski diagrams) in the rest frames of A, B, and C. Students are typically very resistant to drawing diagrams, but I cannot recommend Minkowski diagrams enough. They are how I learned to understand special relativity.

 

[DrGreg]

To expand on Ibix's first sentence:

What will A’s clock read at Event 2?

A's clock is not at Event 2, so the question has no answer.

I guess what you mean is:

• What will A's clock read at the event at A that is simultaneous with Event 2?

But that has no answer, either, because "simultaneous" depends on which inertial frame you use. There are 3 inertial frames in this problem, the rest frame of A, the rest frame of B and the rest frame of C. The 3 frames will give you 3 different answers to your question.

 

[h1a8]

To expand on Ibix's first sentence:

A's clock is not at Event 2, so the question has no answer.

I guess what you mean is:

• What will A's clock read at the event at A that is simultaneous with Event 2?

But that has no answer, either, because "simultaneous" depends on which inertial frame you use. There are 3 inertial frames in this problem, the rest frame of A, the rest frame of B and the rest frame of C. The 3 frames will give you 3 different answers to your question.

Ok I understand that question 1 depends of which inertial frame (each frame gives a different answer).
But I have no clue on how to answer question 2 when C passes A.

 

[Ibix]

But I have no clue on how to answer question 2 when C passes A.

Erm... you already did it. Your first answer of 12.5 years was correct.

Do you mean that you don't understand why your second approach was incorrect? In that case I'd refer you to the third paragraph of my last post. Do you know what the Lorentz transforms are?

 

[h1a8]

Question 1 is ill-posed and has no unique answer. Not understanding this is why you have different answers to question 2. Your first attempt at question 2 is correct, and you failed to account for the relativity of simultaneity in your second attempt. (You probably failed to account for it in your first attempt too, but happened to pick an approach where it has no effect.)

This kind of simple problem can always be analysed by picking an inertial frame (I'd suggest A's rest frame), writing down the coordinates of interesting events in that frame, then using the Lorentz transforms to calculate times and places in other frames. Don't blindly apply the time dilation equation and hope - it's a special case of the Lorentz transforms and practice is needed to see when you can use it.

In this particular case it would be helpful to use the Lorentz transforms to work out what time A's clock shows at the same time as B and C pass in the rest frames of A, B, and C. Have a go - post your working if you get stuck or need help interpreting the answers.

Probably the more helpful solution is to draw spacetime diagrams (also called Minkowski diagrams) in the rest frames of A, B, and C. Students are typically very resistant to drawing diagrams, but I cannot recommend Minkowski diagrams enough. They are how I learned to understand special relativity.

Simple problem? Wow I been at this for months. I feel slow lol.

I have tried everything (looking at it in all frames, including A's).
I don't like diagrams (I don't understand them that well). I rather use math equations.
Using A's frame, I used Lorentz transform to derive that tA = 5/4tC (a speed difference of 3/5c gives that result). That was simple algebra and also a well known result from countless sources. So when A's clock read t = 6.25 then C should be passing by (C's clock reads t = 5).

I can work most basic math out (algebra, Calculus, ordinary DiffyQ, etc). I just need the relevant equations to use (which I do not know I'm assuming). I'll I know is lorentz transform equations.

 

[h1a8]

Erm... you already did it. Your first answer of 12.5 years was correct.

Do you mean that you don't understand why your second approach was incorrect? In that case I'd refer you to the third paragraph of my last post. Do you know what the Lorentz transforms are?

Oh ok. I didn't quite understand you at first. So the first attempt is correct but not the 2nd attempt (for the 2nd question).

So ignoring both event 1 and object B (making a more simpler scenario) and assuming that, according to frame C, it takes t = 5 to pass A after traveling v = 3/5c towards A. This would make A's clock read t = 4 at the time of passing right? Since tC = 5/4tA

 

[PeterDonis]

So when A's clock read t = 6.25 then C should be passing by (C's clock reads t = 5).

You mean when A's clock reads 12.5 then C should be passing by with C's clock reading 10. That's the correct answer you gave in the OP of this thread (your first attempt, item 2).

Using A's frame, I used Lorentz transform to derive that tA = 5/4tC

That's not a full Lorentz transformation, it's just calculating the gamma factor.

For a full Lorentz transformation, you first need to write down the coordinates of all the events of interest in the original frame. If the original frame is A's rest frame, those coordinates would be:

Event 1: (x, t) = (0, - 12.5)

Event 2: (x, t) = (3.75, - 6.25)

(Note: You did not calculate the x coordinate of event 2--i.e., its distance from A in A's rest frame. Can you see how the value I give above was obtained?)

Event 3: (x, t) = (0, 0)

Then you need to use the Lorentz transformation equations to obtain the coordinates of the same events in some other frame, such as C's rest frame. Note that the simplest way to do this is to make the origin of both frames the event where A and C meet, i.e., event 3; that is why the coordinates of that event above are (0, 0) and the time coordinates of events 1 and 2 are negative.

Since you have said you prefer math equations, I suggest that you use the Lorentz transformation equations to obtain the coordinates of the above three events in C's rest frame. Note that the relative velocity $v$ that you must use is negative, i.e., $v = - 3/5$, because C is moving in the negative $x$ direction.

 

[Dragon27]

So ignoring both event 1 and object B (making a more simpler scenario) and assuming that, according to frame C, it takes t = 5 to pass A after traveling v = 3/5c towards A. This would make A's clock read t = 4 at the time of passing right? Since tC = 5/4tA

You forgot to take into the account the initial time on the clock. A's clock would read $t_0+4$, where $t_0$ is the time on the A's clock in C's frame of reference when C's clock starts ticking. And (for nitpicking's sake) in C's frame of reference it is A who's traveling towards C, not C traveling to A.

 

[Sagittarius A-Star]

assuming that, according to frame C, it takes t = 5 to pass A after traveling v = 3/5c towards A.

You should also consider length-contraction and relativity of simultaneity, not only time-dilation.

If you imagine a ruler at rest in A's frame, where event 1 and event 3 happen at one end of the ruler and event 2 happens at the other end of the ruler, then the ruler is length-contracted in C's rest frame.

As others said, if you use the complete Lorentz transformation, then all relevant effects are automatically covered.

• length contraction
• relativity of "same location"
• time dilation
• relativity of "same time"

$\require{color} x' = \color{blue}\gamma \color{black}(x\color{red}-vt\color{black})$
$t'= \color{green} \gamma \color{black}(t \color{orange}-\frac{v}{c^2}x\color{black})$

 

[robphy]

Spacetime diagrams are great for organizing one's thoughts
and giving interpretations to various symbols in formulae.

In the end, it can provide a summary of various calculations and interpretations
that one did [and many more that are suggested .. or could be suggested... by others]
to provide fuller storylines for what happened.

Spoiler: a spacetime diagram (made fancy)

a spacetime diagram (...made fancy with tickmarks shown)

By drawing the grid of C's diamonds (instead of A's as above), you effectively do a Lorentz boost (and translation).

 

[PeroK]

@h1a8 here's a better one. A remains at rest. B moves to the right with some velocity $v$. When A's clock reads $1s$, B's clock reads less than $1s$ - by relevant gamma factor.

C moves to the left at the same speed $v$. When A's clock reads $1s$, C's clock reads the same as B's. These two clocks, therefore, are synchronised.

But, B and C are in relative motion with a relative speed greater than $v$. So, there must be mutual time dilation. The clocks cannot remain synchronized.

Explain!

 

[vanhees71]

Well, let's do the algebra. First we describe all three world lines in A's inertial frame $\Sigma$ and parametrize them in A's proper time, which is the coordinate time of $\Sigma$:
$$x_A=\begin{pmatrix}c \tau_A \\ 0 \end{pmatrix}, \quad x_B=\begin{pmatrix} c \tau_A \\ v \tau_A \end{pmatrix}, \quad x_C=\begin{pmatrix} c \tau_A \\ -v \tau_A \end{pmatrix}.$$
Of course $\tau_B=\sqrt{x_B \cdot x_B}/c=\tau_A/\gamma=\tau_C$.

Now we transform to the rest frame of B, which is done by the Lorentz boost
$$\hat{\Lambda}=\begin{pmatrix} \gamma & -\gamma \beta \\ -\gamma \beta & \gamma \end{pmatrix}.$$
In B's frame $\Sigma'$ the trajectories read
$$x_A'=\begin{pmatrix} c \gamma \tau_A \\ -v \gamma \tau_A\end{pmatrix}, \quad x_B'=\begin{pmatrix} c \tau_A/\gamma \\0\end{pmatrix}, \quad x_C'=\begin{pmatrix}\gamma (1+\beta^2) c \tau_A \\ -2v \gamma \tau_A \end{pmatrix}.$$
That answers the question: The proper times of B and C are the same at any time $\tau_A$, but the coordinate times of B and C in B's rest frame, $\Sigma'$ are not the same at given $\tau_A$ (except of course for $\tau_A=\tau_B=\tau_C=0$, where all three observers are at the same place. Of course, there's nothing paradoxical. One must only distinguish between different notions of time. The only unique physical times are the frame-independent proper times of each observer.

 

[Sagittarius A-Star]

2. Since A is moving at v_A = 3/5c relative to C then t_C=5/4t_A. So, when t_C=10 then t_A=8.
Contradiction?

By calculating backwards from $t_A = 12.5$ at event $E_3$ you can see, that there is no contradiction.

Because object $A$ is moving and therefore time-dilated with reference to the inertial restframe of object $C$, object $A$ needs $\Delta t_A = 4$ elapsed time to move from event $E_{A2C}$ to event $E_3$. Event $E_{A2C}$ is the event on $A$'s worldline, which happens simultaneously to event $E_2$, with reference to the inertial restframe of object $C$.

At event $E_{A2C}$, $A$'s watch showed $t_A = 12.5 - \Delta t_A = 8.5$.

 

[Sagittarius A-Star]

QUESTIONS:
1. What will A’s clock read at Event 2?

Event $E_2$ does not happen on object $A$'s worldline.

• $A$'s clock reads $4$ simultaneously to event $E_2$, with reference to the inertial restframe of object $B$.
• $A$'s clock reads $6.25$ simultaneously to event $E_2$, with reference to the inertial restframe of object $A$.
• $A$'s clock reads $8.5$ simultaneously to event $E_2$, with reference to the inertial restframe of object $C$.

 

[h1a8]

Event $E_2$ does not happen on object $A$'s worldline.

• $A$'s clock reads $4$ simultaneously to event $E_2$, with reference to the inertial restframe of object $B$.
• $A$'s clock reads $6.25$ simultaneously to event $E_2$, with reference to the inertial restframe of object $A$.
• $A$'s clock reads $8.5$ simultaneously to event $E_2$, with reference to the inertial restframe of object $C$.

View attachment 304355

How did you get that A's clock read 8.5 simultaneously to event 2 with reference to C?

 

[Sagittarius A-Star]

How did you get that A's clock read 8.5 simultaneously to event 2 with reference to C?

You can see the answer in my above posting #17.

 

[robphy]

How did you get that A's clock read 8.5 simultaneously to event 2 with reference to C?

You can also look at the diagram (cloaked by the spoiler tag) in #14.

 

[h1a8]

You can also look at the diagram (cloaked by the spoiler tag) in #14.

I don't see the logic of how C's line points to 8.5 on A's line at event 2.

 

[h1a8]

By calculating backwards from $t_A = 12.5$ at event $E_3$ you can see, that there is no contradiction.

Because object $A$ is moving and therefore time-dilated with reference to the inertial restframe of object $C$, object $A$ needs $\Delta t_A = 4$ elapsed time to move from event $E_{A2C}$ to event $E_3$. Event $E_{A2C}$ is the event on $A$'s worldline, which happens simultaneously to event $E_2$, with reference to the inertial restframe of object $C$.

Wouldn't that be circular? Working backwards (assuming A's clock reads 12.5) to obtain results so that there is no contradiction.
Is there a forward process (without graphs and assuming A's clock reads 12.5 at event 3) that tells us A's clock reads 8.5 in C's frame at event 2?

At event $E_{A2C}$, $A$'s watch showed $t_A = 12.5 - \Delta t_A = 8.5$.

 

[Dragon27]

Is there a forward process (without graphs and assuming A's clock reads 12.5 at event 3) that tells us A's clock reads 8.5 in C's frame at event 2?

Do you know how to calculate the coordinates of any event in the frame of reference B starting from its coordinates in the frame of reference A, when B moves with some constant speed relative to A? (hint: the name of the transformation has been mentioned in posts #5 and #8).

 

[Sagittarius A-Star]

Is there a forward process (without graphs and assuming A's clock reads 12.5 at event 3) that tells us A's clock reads 8.5 in C's frame at event 2?

I assume, with "... in C's frame at event 2"
you mean
"... simultaneous to event 2 with reference to the restframe of C".

Here’s my attempt
1. Since B is moving at v_B = 3/5c relative to A then t_A=5/4t_B. So, when t_B=5 then t_A=6.25.

You calculated correctly "forward", that A's clock reads $6.25$ at event $E_{A2A}$, which is simultaneous to event $E_2$ with reference to A's frame.

Let me call A's frame $S$ and C's frame $S'$.

With reference to A's frame I calculate now the $\Delta t$ between event $E_2$ and event $E_{A2C}$, which is simultaneous to event $E_2$ with reference to C's frame ($\require{color} \color{red} \Delta t' = 0 \color{black}$).

The spatial distance $\Delta x'$ between event $E_{A2C}$ and all events on C's worldline (including $E_2$ and $E_3$), with reference to C's frame, is
$\require{color} \Delta x' = 5 \cdot \frac{3}{5} = \color{blue}3\color{black}$.

("Travel distance", Object A needs 5 seconds coordinate-time to move from event $E_{A2C}$ to object C, with reference to C's rest frame.)​

I will use the Lorentz transformation (with unit system $c:=1$).

$\require{color} \Delta t = \gamma (\Delta t' + \frac{v}{c^2} \Delta x') = \frac{5}{4} (\color{red}0\color{black} + \frac{3}{5} \cdot \color{blue}3\color{black}) = 2.25$.

Adding this result to your $6.25$ gives:
$6.25 + 2.25 = 8.5$.

 

[Ibix]

Is there a forward process (without graphs and assuming A's clock reads 12.5 at event 3) that tells us A's clock reads 8.5 in C's frame at event 2?

The Lorentz transforms and the process I suggested in #5. In this case, the explicit process is as follows.

Let event #1 be (0,0) in all frames so we can use the Lorentz transforms easily. Let A's rest frame coordinates be denoted $t_A,x_A$, and similarly for B's and C's rest frames.

Event #2 (crossover) occurs at $t_{B2}=5$ $x_{B2}=0$, where the B2 subscript tells us this is B's description of event 2. You want to know the time A's clock shows at the event simultaneous with this according to C. So use the Lorentz transforms to work out what $t_{C2}$ and $x_{C2}$ are (if you know the velocity composition formula you can find C's velocity according to B and do this directly; otherwise it may be easier to compute the A coordinates and then the C coordinates).

Once you have the coordinates of event 2 according to C (note the time will not be zero - C's shipboard clock is offset from their coordinate time) then you want the event that is simultaneous with this according to C but at A's location. So you know that at this event $x_A=0$ and $t_C=t_{C2}$, and you can use the Lorentz transforms to find the $t_A$ value (and $x_C$ value if you wish).

You already know that the answer must be $t_A=8.5$.

 

[pervect]

I don't see the logic of how C's line points to 8.5 on A's line at event 2.

To appreciate the diagram, there are 2 things you need. The first is the ability to draw, or at least understand, space-time diagrams. To make this more specific, you need to know how to draw the space-time diagram of a light clock. We will give an example below, but first we will talk about the second thing you need to know, which is the Einstein clock synchronization convention. This convention formally defines what we mean by clocks being syncrhonized in a given frame. We will now use a space-time diagram to explain the Einstein convention.

First let us focus on the red square.

The red square, with events ABCD is a light clock.

AB represents a light signal going to the left emitted from A and going to B. BC represents the light beam AB continuing on, after it is reflected at B, now going to the right.

AD represents a light signal starting at A and going to D, and DC represents the previous light signal reflected and going back to the left.

On this diagram, AC (not drawn) is a timelike worldline, that represents a clock traveling from event A to event C. The line segment BD (not drawn) is a spacelike worldine, representing two different events, B and D, that occur at "the same time" in the frame in which AC represents the worldline of a clock at rest.

If we consider some event O in the middle of the spacelike worldine BD, we can say that events B, O, and D all occur at the same time. The reasoning for this is that events B and D are equally distant from the worldline AC, and that because light travels at a constant speed, the time it takes the light signal to go from A to B is the same amount of time that it takes for the light signal to go from A to C.

This is all simple enough, but where things may get harder to understand is that EFGH also represents a light clock. The logic is the same for the light clock ABCD, and we can conclude that in the frame in which EG is at rest (which is not the frame the diagram is drawn in, unlike the previous case) that events F and H must be simultaneous.

Given that one understands WHY events B and D are simultaneous in the frame in which AC is at rest, and that one understand WHY the events F and H are simultaneous in the frame in which EG is at rest should explain Rob's diagram.

So, the two things one needs to understand are how to draw and interpret space-time diagrams, and the Einstein definition of what simujltaneity means in some specific frame. A related point one needs to understand is that in special relativity, simultaneity is FRAME DEPENDENT. Two events that are simultaneous in one frame are NOT simultaneous in another frame. In our diagram, events B and D are simultaneous ONLY in the frame in which AC is at rest, they are not simultaneous in the frame in which EG is at rest. Similarly, events F and H are simultaneous ONLY in the frame in which EG is at rest, they are NOT simultaneous in a frame in which AC is at rest.

There are some more advanced points to be made, specifically the significance of "equal areas" of the light clocks ABCD and EFGH being the same, but that's probably left for some future date. The important thing to grasp is the fact that in special relativity, simultaneity is FRAME DEPENDENT, and to understand the operational definition of simultaneity using light clocks.

 

[Sagittarius A-Star]

Note that the simplest way to do this is to make the origin of both frames the event where A and C meet, i.e., event 3

Let event #1 be (0,0) in all frames so we can use the Lorentz transforms easily.

There exists also the possibility to not specify an event as common origin of the frames. If only differences between coordinates are transformed (for example $\Delta x$ instead of $x$), then the offsets to the respective origin of the axes cancel out. I used this in posting #25. I think this is the easiest way, if you are only interested in coordinate differences and if you have more than two objects/frames.

 

[PeterDonis]

There exists also the possibility to not specify an event as common origin of the frames.

Not really. What you are doing is mathematically equivalent to performing a translation to make the spacetime origins of the frames coincide, but the language you use to describe this obfuscates what is actually going on. "Coordinate differences" is just another name for "coordinates when the origin is shifted by a translation". They're not different things.

 

[Sagittarius A-Star]

"Coordinate differences" is just another name for "coordinates when the origin is shifted by a translation". They're not different things.

Let's assume 2 frames $S$ and $S'$. Define 2 events, which have in frame $S$ the coordinates
$E_1: (ct, x) = (0, 1)$
$E_2: (ct, x) = (0, 2)$

As example, frame $S'$ shall have as origin event $E_1$ (= different from the origin of frame $S$). Related LT for x is:
$x' = \gamma ((x-1) -vt)$

Transformation of the spatial distance between both event:
$\require{color}\Delta x' := x'_2 - x'_1 = \gamma ((x_2-1) -vt_2) - \color{blue} \gamma ((x_1-1) -vt_1) \color{black}= \gamma ((x_2 - x_1) -1 +1 - v(t_2 - t_1))$
$\Rightarrow$
$\Delta x' = \gamma (\Delta x - v \Delta t) \ \ \ \ \$(1)

Equation (1) would also be valid, if I had chosen the same event as origin for both frames.

 

[PeterDonis]

Equation (1) would also be valid, if I had chosen the same event as origin for both frames.

Yes. This is in agreement with what I said. Just remove the $\Delta$s from Equation (1) and you have the Lorentz transformation bewteen the two frames when they both have the same event as the origin.

 

[Sagittarius A-Star]

Just remove the $\Delta$s from Equation (1) and you have the Lorentz transformation bewteen the two frames when they both have the same event as the origin.

Yes. If $\Delta x$ gets replaced by $x$ (and the same for y, z, t coordinates) , this means, that $x_2 - x_1$ gets replaced by $x - 0$. This would be equivalent to defining the origin as one of the two events.

Source for Lorentz transformation of deltas and differentials (equations 7 and 8):
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

 

[vanhees71]

Of course, Minkowski space is not only a Lorentzian vector space but an affine manifold with the translations as additional symmetry, i.e., the complete continuous symmetry group is the proper orthochronous Poincare group, including space-time translations.